| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Conditional divisibility with if-then |
| Difficulty | Challenging +1.2 Part (a) requires completing a standard proof by contradiction with most work already done—students just need to recognize the missing case p=3k+2 and state the contradiction. Part (b) is a direct application of part (a) using the rational root template (assume √³3 = a/b, derive contradiction). While proof questions can intimidate students, this is highly scaffolded with clear structure and familiar techniques from the A-level proof syllabus. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Let \(p = 3k+2\), then \((3k+2)^3 = 27k^3 + 54k^2 + 36k + 8\) | M1 | Attempts to expand \((3k+2)^3\) or \((3k-1)^3\); cubic with at least two correct terms (allowing incorrect signs) |
| \(= 3\times(9k^3 + 18k^2 + 12k + 3) - 1\), not a multiple of 3 | A1 | Achieves correct \(3\times(\ldots)+r\), \( |
| So \(p\) cannot be of form \(3k+1\) or \(3k+2\), since \(p^3\) is a multiple of 3. Hence \(p\) must be a multiple of 3, a contradiction. Therefore for all integers \(p\), when \(p^3\) is a multiple of 3, \(p\) is a multiple of 3. | A1 | Must reference both cases, give a "contradiction" and indicate proof is complete |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Assumption: there exist integers \(p\) and \(q\) such that \(\sqrt[3]{3} = \frac{p}{q}\) (where \(p\) and \(q\) have no non-trivial common factors) | B1 | Sets up algebraic initial statement to be contradicted; no requirement to state \(p,q\) have no common factors for this mark |
| \(\sqrt[3]{3} = \frac{p}{q} \Rightarrow p^3 = 3q^3\) | M1 | Cubes correctly and multiplies through by \(q^3\) |
| So \(p^3\) is a multiple of 3 and so \(p\) is a multiple of 3 | A1 | Deduces \(p^3\) multiple of 3 hence \(p\) multiple of 3; jumping directly to \(p\) multiple of 3 is A0 |
| But \(p = 3k \Rightarrow 27k^3 = 3q^3 \Rightarrow q^3 = 9k^3\) | dM1 | Sets \(p=3k\) and proceeds to find \(q^3\) in terms of \(k\) |
| Hence \(q^3\) is a multiple of 3 so \(q\) is a multiple of 3, but as \(p\) and \(q\) have no non-trivial common factors, this is a contradiction. Hence \(\sqrt[3]{3}\) is an irrational number.* | A1* | Requires: correct algebra, correct deductions in order, initial statement included \(p,q\) have no common factors, correct contradiction and conclusion, no contrary statements during proof |
## Question 9(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Let $p = 3k+2$, then $(3k+2)^3 = 27k^3 + 54k^2 + 36k + 8$ | **M1** | Attempts to expand $(3k+2)^3$ or $(3k-1)^3$; cubic with at least **two correct** terms (allowing incorrect signs) |
| $= 3\times(9k^3 + 18k^2 + 12k + 3) - 1$, not a multiple of 3 | **A1** | Achieves correct $3\times(\ldots)+r$, $|r|<10$ form and states not a multiple of 3 |
| So $p$ cannot be of form $3k+1$ or $3k+2$, since $p^3$ is a multiple of 3. Hence $p$ must be a multiple of 3, a contradiction. Therefore for all integers $p$, when $p^3$ is a multiple of 3, $p$ is a multiple of 3. | **A1** | Must reference **both cases**, give a "contradiction" and indicate proof is complete |
## Question 9(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Assumption: there exist integers $p$ and $q$ such that $\sqrt[3]{3} = \frac{p}{q}$ (where $p$ and $q$ have no non-trivial common factors) | **B1** | Sets up algebraic initial statement to be contradicted; no requirement to state $p,q$ have no common factors for this mark |
| $\sqrt[3]{3} = \frac{p}{q} \Rightarrow p^3 = 3q^3$ | **M1** | Cubes correctly and multiplies through by $q^3$ |
| So $p^3$ is a multiple of 3 and so $p$ is a multiple of 3 | **A1** | Deduces $p^3$ multiple of 3 hence $p$ multiple of 3; jumping directly to $p$ multiple of 3 is A0 |
| But $p = 3k \Rightarrow 27k^3 = 3q^3 \Rightarrow q^3 = 9k^3$ | **dM1** | Sets $p=3k$ and proceeds to find $q^3$ in terms of $k$ |
| Hence $q^3$ is a multiple of 3 so $q$ is a multiple of 3, but as $p$ and $q$ have no non-trivial common factors, this is a contradiction. Hence $\sqrt[3]{3}$ is an irrational number.* | **A1*** | Requires: correct algebra, correct deductions in order, initial statement included $p,q$ have no common factors, correct contradiction and conclusion, no contrary statements during proof |\begin{enumerate}
\item A student was asked to prove, for $p \in \mathbb { N }$, that\\
"if $p ^ { 3 }$ is a multiple of 3 , then $p$ must be a multiple of 3 "
\end{enumerate}
The start of the student's proof by contradiction is shown in the box below.
Assumption:\\
There exists a number $p , p \in \mathbb { N }$, such that $p ^ { 3 }$ is a multiple of 3 , and $p$ is NOT a multiple of 3
Let $p = 3 k + 1 , k \in \mathbb { N }$.
$$\text { Consider } \begin{aligned}
p ^ { 3 } = ( 3 k + 1 ) ^ { 3 } & = 27 k ^ { 3 } + 27 k ^ { 2 } + 9 k + 1 \\
& = 3 \left( 9 k ^ { 3 } + 9 k ^ { 2 } + 3 k \right) + 1 \quad \text { which is not a multiple of } 3
\end{aligned}$$
(a) Show the calculations and statements that are required to complete the proof.\\
(b) Hence prove, by contradiction, that $\sqrt [ 3 ] { 3 }$ is an irrational number.
\hfill \mbox{\textit{Edexcel P4 2023 Q9 [8]}}